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Re: (AB)^2 + (BD)^2 [#permalink]
What a Trap!
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Re: (AB)^2 + (BD)^2 [#permalink]
this is so direct yet so confusing. we don't know its a rectangle. hence D
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Re: (AB)^2 + (BD)^2 [#permalink]
So sneaky!!
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Re: (AB)^2 + (BD)^2 [#permalink]
e have a rectangle.
By Pythagoras theorem, Diagonal AD2=AB2+BD2AD2=AB2+BD2
Therefore, we can replace qty A with AD2AD2
The main idea the question is trying to test is that side AB and BD could both be an integer number or a fractional number.
In case ABAB and BDBD are fractional number their square will be less and hence ADAD would be bigger than AD2AD2. However if ABAB and BDBD are integer number AD2AD2 will be bigger than ADAD. Also keep in mind that ABAB and BDBD cannot be 00 or −ve−ve number.
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Re: (AB)^2 + (BD)^2 [#permalink]
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